S. from Utah asks, "Why do math teachers make us simplify radical expressions?"

Great question, S! The good news is, we have a unit on "Simplifying Radical Expressions" in our reference section, and the unit has a page that answers your question! If you visit this page right here: Why We Simplify Radicals, you'll get an explanation for some of the reasons.

The short answer is: There are multiple ways of writing the same number, but if we all follow the simplification rules, we'll all write our answers the same way, which makes it easier for people to work together, if they're all expressing their work in the same way.

Here's another way of looking at it: suppose you lived at 12 Main Street. Which of the following pieces of mail are addressed to you?

• 12 Main St
• SQR(144) Main St
• 2SQR(36) Main St
• 4SQR(9) Main St
• 3SQR(16) Main St
• SQR(9)SQR(16) Main St

Which of these pieces of mail are addressed to you? Well, obviously the first one is, but even though the others might not look like it, they all actually evaluate to 12 Main St. But the big question is: how many of them will actually get delivered to you? The answer is probably, "It depends on how much of a sense of humor the postal workers have!"

The reality is, if we don't simplify our answers so they all look the same, we make a lot of extra work for someone. Right now, that someone is your teacher (who has to grade your tests and homework), but if you pursue a career in a STEM field, the someone will be you, and the people on your team.

So practice simplifying those radicals until it becomes second nature. Your teacher will thank you now, and someday future-you may thank you as well!

Long-time puzzler and player Dan from Minnesota asks the following question about our Entrapment Game:

My question is about the strategy game called Entrapment . After playing and solving several of these puzzles over the years I couldn't help notice the occurrence of very difficult and most brilliant puzzles that can take me hours to solve, verses only seconds to solve others, My question is are these randomly generated or are some these more advanced puzzles the result of mischievous talented geniuses? 

Hi Dan, believe it or not, all of these games were the result of just one mischief maker. Me. Or, more precisely, my programming. There are almost 7,000 entrapment puzzles on the site, and they were all built by the software using an algorithm I built to create the puzzles at random.

How does it work? Well, the software starts by picking a random location for one of the gray dots, and then randomly picking a location for one of the red dots. Having picked one location, it now knows the location for a second red dot (since it has to be placed in such a way that the gray dot is midway between the two red dots). The software continues on in this way, alternately picking locations for gray dots and red dots, with the proviso that if the software can find a place to put a gray dot which is halfway between two gray dots already on the board, it skips putting another red dot on the board. In this way, the number of gray dots grows larger than the number of red dots.

In the end, the gray dots are removed from the board, leaving just the red ones. Of course, there's no guarantee that there's only one solution, and you've probably noticed that in some cases there are more red dots than are needed. This is because you found a solution that is different than the one the computer found.

Because the process is random, some puzzles end up being more challenging than others, and there's no good way to programmatically evaluate which puzzles are most challenging. Currently we're tracking the average amount of time taken for each puzzle, so that eventually we can break the puzzles into easy/hard/expert categories. But, of course, with 7,000 puzzles on the site, it's going to take a long time to get enough plays on each puzzle that we can confidently categorize them.

In the meantime, thank you for playing, and helping to add to our difficulty calculations!

A visitor for the US asks the following question: "Who is the father of light?"

"Father of Lights" (note that I'm using the plural of "light") is a phrase which is found in the Bible, and is used to describe God. The reference, in case you want to know, is James 1:17, which reads as follows (using the King James translation):

Every good gift and every perfect gift is from above, and cometh down from the Father of lights, with whom is no variableness, neither shadow of turning.

Reading this in a modern translation (New International Version):

Every good and perfect gift is from above, coming down from the Father of the heavenly lights, who does not change like shifting shadows.

What does the phrase mean? Well, that's a good question! And there are a few ways to take meaning from the title.

1. It is a reference to creation; God as the creator of all the heavenly lights, is referred to as the father of them. In a sense, then, you could give Him the title "Star Father."
2. It is a reference to Jesus, who called himself "the light of the world." Following the logic of that; if Jesus is the light, and God is his father, then God is the "Father of Light."
3. Light, in the Bible, is usually associated with either knowledge or righteousness (or both). For example, Isaiah 9:2 says that "the people who walked in darkness have seen a great light," and according to one Bible dictionary, the word "darkness" there metaphorically means one of the following: misery, destruction, death, ignorance, sorrow, or wickedness. Thus, of course, light would be, metaphorically, the opposite. So saying "Father of Light" could be the equivalent of "Father of Knowledge" or "Father of Righteousness."

The conclusion of the verse, which says he is unchanging, and has no shifting shadows, may be a reference to the stars, which twinkle and shift - indicating that God is greater than the stars of the heavens.

Thanks for asking, and I hope that was helpful!

Professor Puzzler

Judah asks: "A Highly Composite number is a number that has more factors than any other less than it (positive, whole numbers). Like 12 has 6 factors, which is more than any other below it (whereas 1 has 1 , 2 has 2, and so on). Is there an equation to tell if a number is highly composite? If no, than I might find one. :P But is there?"

Hi Judah,

I'm not familiar with an actual formula that you could use (something along the lines of a function that you plug n into it and you get the n^th highly composite number out). However, there are some tips I can give you for finding highly composite numbers (short of looking up a list on a website somewhere).

First, it's important to note that what really matters in highly composite numbers is the exponents in their prime factorization. Why? Because the number of factors a number has can be directly calculated from those exponents. Let me show you how.

Take the number 72. How many factors does this number have? We can find out by simply testing numbers to find out if they're factors:

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

72 has twelve factors.

But there's another way to find out how many factors 72 has. Write the prime factorization of 72:

72 = 2³3²

Now consider the fact that every factor of 72 contains both powers of 2 and powers of 3. For example, 12 = 2²3¹, 18 = 2·3². This is also true of 1, though you might not think so at first: 1 = 2⁰3⁰. Similiarly, 8 = 2³3⁰.

But, of course, if a number has 2⁴ or 3³ as part of its prime factoriztion, the number can't be a factor of 72, because it has too many twos, or too many threes.

We can look at it this way: if a number is a factor of 72, it contains in its prime factorization either 2⁰, 2¹, 2², or 2³. Similarly, the prime factorization contains 3⁰, 3¹, or 3².

There are 4 ways to choose how many twos to include, and 3 ways to choose how many threes to include to build a factor of 72. If you've studied counting principles at all, you know that the total number of ways to create a factor is then 4 x 3 = 12, which matches the number we obtained by listing factors.

Let's try another example: How many factors does 6,615 have? I'm sure you don't want to try to list them all by hand, right? So let's try our other method. Prime factorization of 6615 = 3³5¹7².

Thus, there are 4 ways to choose how many threes, 2 ways to choose how many fives, and 3 ways to choose how many sevens to include in a factor. Therefore, there are 4 x 2 x 3 = 24 factors.

So how do we use this to help us find highly composite numbers? Well, we can use it indirectly, because we can use this to help find the smallest number that has n factors.

Let's say we want to find the smallest number with 15 factors. Well, 15 can be factored in two ways:

15 = 15 x 1
15 = 3 x 5

Using the first factorization is going to result in a ridiculously large number (2¹⁴3⁰), so should use the other way of factoring. We need to have a prime factorization that includes an exponent of 2 and an exponent of 4 (2 is one less than 3, and 4 is one less than 5). We'll get our smallest number by assigning the largest exponent to the smallest prime factor, which gives us 2⁴3² =  144. 144 is the smallest number with 15 factors.

What if we wanted to find the smallest number with 16 factors? There are many more possibilities here, because 16 can be factored in several different ways:

16 = 16 x 1
16 = 8 x 2
16 = 4 x 4
16 = 4 x 2 x 2
16 = 2 x 2 x 2 x 2

Each of these gives rise to a different number:

16 = 16 x 1            21530 = 32768
16 = 8 x 2             2731 = 384
16 = 4 x 4             2333 = 216
16 = 4 x 2 x 2         233151 = 120
16 = 2 x 2 x 2 x 2     21315171 = 210

Of these 120 is the smallest, so 120 is the smallest number with 16 factors.

Notice that the smallest number with 16 factors is actually smaller than the smallest number with 15 factors, which means there is no highly composite number with 15 factors. You can't assume that 120 is a highly composite number, either, until you've done some more testing, but if I had to make a guess, I'd say it probably is. Okay, I just checked a list on Wikipedia, and 120 is a highly composite number.

Hopefully this will inspire you to more exploration, and maybe you'll find that formula you were looking for. Thanks for asking!

Esther from Georgia wants to know "Why do extraneous solutions happen in math problems?"

Extraneous solutions are solutions which are obtained algebraically, but even though they are found algebraically, they don't work in the context of the problem. Here are some of the reasons a problem might have extraneous solutions. For each reason, I'll give an example.

Real World Restrictions

Sometimes a solution that we find algebraically is extraneous because it doesn't fit real world restrictions on the solution. For example, consider the following problem:

A rectangle's length is 2 more than its width. Its area is 48 square feet. What is the rectangle's width?

To solve this, we say, let w equal the width. Then we have: w(w + 2) = 48

This is a quadratic equation: w² + 2w - 48 = 0, which leads to w = 6 or w = -8. However, the real world restriction on the solution is that w is the width of a rectangle, and rectangles can't have negative side lengths. Therefore, -8 is an extraneous solution.

Problem Restrictions

In other cases, the problem itself might set restrictions on the solution.

Find all two digit numbers such that when the number is squared, and ten times the number is subtracted, the result is 11.

This problem leads to the quadratic: x² - 10x - 11 = 0

(x - 11)(x + 1) = 0

x = 11 or x = -1.

But the problem itself sets the restriction that x must be a two digit number, so x = -1 is extraneous.

Division By Zero

Sometimes a problem has fractions, and a number that works out algebraically actually causes a division by zero, which is illegal, immoral, and socially unacceptable. As a problem writer for math competitions, I like to deliberately create problems in which this happens, because the very best students check their solutions and realize they've got an extraneous solution. Here's a very simple example:

(x + 1)/(x - 1) = 2/(x - 1)

An easy way to solve this is to multiply both sides by (x - 1), which gives x + 1 = 2, or x = 1. But the process of multiplying by (x - 1) has made a fundamental change to the problem; we no longer have a denominator. If we go back to the original problem and try to plug x = 1 into it, we find that we have a division by zero, and therefore x = 1 doesn't actually work; it's extraneous.

Principle Roots

Each positive number has two real square roots. For example, the square roots of 4 are 2 and -2. However, when we write radical notation, we are, be definition, referring to the principle square root, which is the positive value. Thus, a solution may be extraneous because it results from using a negative square root instead of the principle square root. Here's an example:

√(x)  = x - 6

To solve this, we square both sides:

x = x² - 12x + 36

Rearrange, simplify, factor, solve:

x² - 13x + 36 = 0

(x - 9)(x - 4) = 0

x = 9 or x = 4

Now, plug 9 into the original equation, and you'll see that it does work. But what happens when you plug in 4? You get a false equation! But why does that happen? Because on the left-hand side, 4 has two square roots, and one of them (-2) does work. So it's our own choice to define √(x) as the positive square root of x that results in the extraneous solution.

Are there other causes for extraneous solutions? Probably, but that should serve as a simple crash course on the common ones you'll see in an algebra class.